I am trying to understand the definition of frechet distance but I am struggling to understand the reparameterization part in the definition. I got the following definition from wikipedia

Let A and B be two given curves in S. Then, the Fréchet distance between $\mathbf A$ and $\mathbf B$ is defined as the infimum over all reparameterizations $\alpha$ and $\beta$ of $[0,1]$ of the maximum over all $t \in [0,1]$ of the distance in $\mathbf S$ between $\mathbf{A(\alpha(t))}$ and $\mathbf{B(\beta(t))}$.

What does it mean by reparameterizations $\alpha$ and $\beta$ of $[0,1]$ of the maximum over all $t \in [0,1]$? Can anyone explain it with an example.

1 Answer
1

The wikipedia page (https://en.wikipedia.org/wiki/Fréchet_distance ) refers to a reparametrization as a continuous, non-decreasing surjection, so the Fréchet distance considers all possible curves $\alpha$ and $\beta$ that are non-decreasing surjections, finds the value of $t$ at which their difference is a maximum, and then takes the greatest of all those maxima.

A surjection from $[0,1]$ to $[0,1]$ is a map $\alpha$ such that there exists some $x$ in $[0,1]$ such that $\alpha(x)=y$ has a solution for all $y\in[0,1]$. It's possible for any $x$ to map to more than one $y$, the important thing is that no $y$ is missed out.

The simplest example of a reparametrization is the trivial one $\alpha_1(x)=x$. This is clearly surjective since $\alpha_1(x)=y$ has the solution $x=y$ for all $y\in[0,1]$. It is also non-decreasing since if $x_1<x_2$ then $\alpha_1(x_1)<\alpha_1(x_2)$ for all $x_1,x_2\in[0,1]$.

A more interesting reparametrization is
$$\alpha_2(x) = \frac{\sin x}{\sin 1}$$ The function $\sin$ is monotonically increasing on $[0,1]$, as is its inverse $\sin^{-1}$ -- see the graphs.

Dividing through by $\sin 1$ ensures that $\alpha(0)=0$ and $\alpha(1)=1$ so we achieve surjectivity on $[0,1]$ as required.

Suppose now you have two parabolic curves: $A(t):= t^2$ and $B(t) = t^2+t/2+1/4$. Reparametrizing with $\alpha_1$ doesn't change the curves at all, but when we reparametrize with $\alpha_2$ we get
$$A(\alpha(t)) = \left( \frac{\sin t}{\sin 1} \right)^2 \quad \mbox{and}\quad B(\alpha(t)) = \left( \frac{\sin t}{\sin 1} \right)^2 + \frac{\sin t}{2 \sin 1} +1/4 $$
Since the parabolae are strictly increasing, and the reparamatrization (being non-decreasing) doesn't change that, they have their maxima at $t=1$, and you can quickly calculate that $|A(1)-B(1)| = \frac{3}{4} = |A(\alpha(1)) - B(\alpha(1))|$